Submission #523774

#	Time	Username	Problem	Language	Result	Execution time	Memory
523774		boykut	K blocks (IZhO14_blocks)	C++14	53 / 100	1104 ms	131356 KiB

blocks

#include <bits/stdc++.h>
 
using namespace std;
 
class sparse_table_min {
public:
	sparse_table_min(vector<long long>&a) : n(a.size()) {
		lg.resize(1 + n);
		lg[0] = lg[1] = 0;
		for (int i = 2; i <= n; i++)
			lg[i] = lg[i / 2] + 1;

		mn.assign(n, vector<long long>(1 + lg[n]));
		for (int j = 0; j <= lg[n]; j++)
			for (int i = 0; i <= n - (1 << j); i++)
				if (j == 0) mn[i][j] = a[i];
				else mn[i][j] = min(mn[i][j-1], mn[i+(1<<(j-1))][j-1]);
	}
	long long query(int l, int r) {
		int d = lg[r-l+1];
		return min(mn[l][d], mn[r-(1<<d)+1][d]);
	}
private:
	int n;
	vector<int> lg;
	vector<vector<long long>> mn;
};

class sparse_table_max {
public:
	sparse_table_max(vector<long long>&a) : n(a.size()) {
		lg.resize(1 + n);
		lg[0] = lg[1] = 0;
		for (int i = 2; i <= n; i++)
			lg[i] = lg[i / 2] + 1;
 
		mx.assign(n, vector<long long>(1 + lg[n]));
		for (int j = 0; j <= lg[n]; j++)
			for (int i = 0; i <= n - (1 << j); i++)
				if (j == 0) mx[i][j] = a[i];
				else mx[i][j] = max(mx[i][j-1], mx[i+(1<<(j-1))][j-1]);
	}
	long long query(int l, int r) {
		int d = lg[r-l+1];
		return max(mx[l][d], mx[r-(1<<d)+1][d]);
	}
private:
	int n;
	vector<int> lg;
	vector<vector<long long>> mx;
};
 
 
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
 
	int n, k;
	cin >> n >> k;

	vector<long long> a(1 + n, 1e18);
	vector<vector<long long>> dp(1 + k, vector<long long>(1 + n, 1e18));

	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}

	sparse_table_max mx(a);
	for (int i = 1; i <= n; i++) {
		dp[1][i] = mx.query(1, i);
	}
	sparse_table_min mn(dp[1]);

 	vector<int> left(n + 1, 0);
	for (int i = 2; i <= n; i++) {
		int l = 1, r = i;
		while (l < r) {
			int m = (l + r) / 2;
			if (mx.query(m, i) == a[i])
				r = m;
			else
				l = m + 1;
		}
		left[i] = r;
	}
	int K = k;
	for (int k = 2; k <= K; k++) {
		for (int i = 2; i <= n; i++) {
			dp[k][i] = min(dp[k][i], dp[k][left[i] - 1]);
			if (mx.query(left[i], i) == a[i]) {
				dp[k][i] = min(dp[k][i], mn.query(left[i] - 1, i - 1) + a[i]);
			}
		}
		mn = sparse_table_min(dp[k]);
	}
 
	cout << dp[k][n] << '\n';
 
	return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 18 / 18
• Subtask 3: 21 / 21
• Subtask 4: 0 / 47